关于c++:Leetcode专题233数字1的个数

力扣链接：
https://leetcode.cn/problems/...
解题思路：

1. 找法则
   从个位数开始，依照以后位将数字分成右边high局部，左边low局部，以后位的数字分为三种类型：
   （1）0: res = high * digit;
   （2）1: res = high * digit + low + 1;
   （3）2/.../9: res = (high + 1) * digit;

class Solution{  public:      int countDigitOne(int n) {          // 参数判断          if (n < 1) {              return 0;          }          // 从个位数开始          long digit = 1;          int high = n / 10, low = 0, cur = n % 10;          int res = 0;          while(cur != 0 || high != 0) {              if (cur == 0) {                  res += high * digit;              }              else if (cur == 1) {                  res += high * digit + low + 1;              }              else {                  res += (high + 1) * digit;              }              low += cur * digit;              high = high / 10;              cur = high % 10;              digit *= 10;          }          return res;      }};